20 mL of 0.1 M acetic acid is mixed with 50 | vedclass

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(B) Let the concentration of potassium acetate be $x \ M$. The mixture forms an acidic buffer. Using the Henderson-Hasselbalch equation: $pH = pK_a +

Vedclass · Accepted Answer

$0.04$

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(B) Let the concentration of potassium acetate be $x \ M$. The mixture forms an acidic buffer. Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[	ext{salt}]}{[	ext{acid}]}$ Given: $pH = 4.8$, $K_a = 1.8 	imes 10^{-5}$, $pK_a = -\log(1.8 	imes 10^{-5}) \approx 4.74$. The number of moles of acid $= 20 \ mL 	imes 0.1 \ M = 2 \ mmol$. The number of moles of salt $= 50 \ mL 	imes x \ M = 50x \ mmol$. Substituting these into the equation: $4.8 = 4.74 + \log \frac{50x}{2}$ $0.06 = \log(25x)$ $25x = 10^{0.06} \approx 1.148$ $x = \frac{1.148}{25} \approx 0.0459 \ M \approx 0.04 \ M$ (rounding to the nearest option).

$20 \ mL$ of $0.1 \ M$ acetic acid is mixed with $50 \ mL$ of potassium acetate. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$ at $27^{\circ} C$. Calculate the concentration of potassium acetate if the $pH$ of the mixture is $4.8$. (in $M$)

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